question_answer

Charges +q and -q are placed at points A and B respectively which are a distance 2Z, apart, C is the midpoint between A and B. The work done in moving a charge +Q along the semicircle CRD is

A)  \[\frac{qQ}{4\pi {{\varepsilon }_{0}}L}\]

B)  \[\frac{qQ}{2\pi {{\varepsilon }_{0}}L}\]

C)  \[\frac{qQ}{6\pi {{\varepsilon }_{0}}L}\]

D)  \[\frac{qQ}{6\pi {{\varepsilon }_{0}}L}\]

Correct Answer: D

Solution :

 Work done is equal to change in potential energy. In 1st case, when charge +Q is situated at C. Electric potential energy of system \[{{U}_{1}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(q)(-q)}{2L}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(-q)Q}{L}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{qQ}{L}\] In IInd case, when charge +Q is moved from C to D. Electric potential energy of system in that case \[{{U}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{(q)(-q)}{2L}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{qQ}{3L}\] \[+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{(-q)(Q)}{L}\] Work done \[=\Delta U={{U}_{2}}-{{U}_{1}}\] \[=\left( -\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{2L}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{qQ}{3L}-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{qQ}{L} \right)\] \[=\left( -\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{2L}-\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{qQ}{L}+\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{qQ}{L} \right)\] \[=\frac{qQ}{4\pi {{\varepsilon }_{0}}}.\left[ \frac{1}{3L}-\frac{1}{L} \right]=\frac{qQ}{4\pi {{\varepsilon }_{0}}}\frac{(1-3)}{3L}\] \[=\frac{-2qQ}{12\pi {{\varepsilon }_{0}}L}=-\frac{qQ}{6\pi {{\varepsilon }_{0}}L}\]